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  1. 1. Nuclear Physics AHL 13.2
  2. 3. Rutherford’s Experiment <ul><li>In 1909 , Rutherford investigated high speed alpha particles that passed through thin, metal foils. </li></ul><ul><li>The experiments were actually carried out by two young physicists named </li></ul><ul><ul><li>Hans Geiger and Ernest Marsden. </li></ul></ul>
  3. 4. Rutherford’s Experiment <ul><li>Alpha particles are double charged helium ions ; </li></ul><ul><ul><li>two protons and neutrons . </li></ul></ul><ul><li>E mitted spontaneously from heavy radioactive atoms such as uranium and plutonium. </li></ul>
  4. 5. Rutherford’s Experiment <ul><li>The beam of alpha particles were directed at a metal foil , detector pick s up the emerging particles on the other side. </li></ul><ul><li>This would give an indication of the nature of the atom. </li></ul>
  5. 6. Rutherford’s Experiment
  6. 7. Rutherford’s Experiment <ul><li>A lead block holds the alpha particle source and emitted through a fine hole in the lead. </li></ul><ul><li>It is collimated into a beam by aligning holes in a series of lead plates. </li></ul><ul><li>The beam is then aimed at a gold foil. </li></ul><ul><li>Gold is used as it can be beaten into a thin foil , about 1  m (8000 atoms) thick. </li></ul>
  7. 8. Rutherford’s Experiment <ul><li>The particles are detected by using a microscope fitted with , a zinc sulfide crystal. </li></ul><ul><li>If an alpha particle hits the crystal, a flash can be seen and counted. </li></ul>
  8. 9. Rutherford’s Experiment <ul><li>Rutherford made predictions as to what he would find. </li></ul><ul><li>He said there would be very few particles would suffer deflections those which did would be deviated , only a few degrees. </li></ul><ul><li>Theory suggested that only 1% would be deflected more than 3 o . </li></ul>
  9. 10. Rutherford’s Experiment <ul><li>Rutherford also determined there was almost no way there could be any backscattering ie. angles greater than 90 o . </li></ul><ul><li>It could only happen with a direct collision with an electron and only 1 particle in 10 3500 would take this path. </li></ul>
  10. 11. Rutherford’s Experiment <ul><li>The calculations were on the basis of previous model of the atom by Thomson. </li></ul><ul><li>Rutherford predicted that the deflections would be due to electrostatic attraction . </li></ul>
  11. 12. Rutherford’s Experiment <ul><li>The actual results came as a great surprise everyone was expecting Thomson’s model was correct. </li></ul><ul><li>One particle in 10 000 was backscattered compared to the theory of 1 in 10 3500 . </li></ul>
  12. 13. Rutherford’s Atom <ul><li>He suggested that all the positive matter was concentrated in </li></ul><ul><ul><li>a very small sphere , </li></ul></ul><ul><ul><li>at the centre of the atom. </li></ul></ul><ul><li>As electrons have very little mass </li></ul><ul><ul><li>all the mass must be in this sphere , called the nucleus. </li></ul></ul><ul><li>Around the nucleus is essentially empty space. </li></ul>
  13. 14. Rutherford’s Atom <ul><li>This explains the backscattering found in the experiment. </li></ul><ul><li>The alpha particle would strongly be repelled when it is directed at the nucleus , due to electrostatic repulsion. </li></ul><ul><li>Those particles that do not hit the nucleus ; would pass through , with little or no deflection. </li></ul>
  14. 15. Rutherford’s Atom
  15. 16. <ul><li>The kinetic energy of the alpha particle is E K </li></ul><ul><li>This kinetic energy is converted into potential energy as the alpha particle overcomes the coulombic force as it approaches the nucleus </li></ul><ul><li>All the kinetic energy = potential energy when the approach is the closest </li></ul>
  16. 17. <ul><li>The potential energy near a point charge is </li></ul><ul><li>E p = Q 1 Q 2 / 4  0 r </li></ul><ul><li>E K = E p = Ze 2e/ 4  0 d </li></ul><ul><li>E K = Ze 2 / 2  0 d </li></ul><ul><li>Calculate the distance of closest approach for an alpha particle of 5MeV approaching a gold nucleus </li></ul>
  17. 19. Nuclear Physics <ul><li>Modern scattering techniques show that the nucleus is not hard like a billiard ball </li></ul><ul><ul><li>but has a ‘fuzzy’ surface. </li></ul></ul><ul><li>This is due to a variation in the density of the nuclear material in the outside layer of the nucleus. </li></ul>
  18. 20. Nuclear Physics <ul><li>The mass of an atom is one of the characteristic properties that can give an insight into the structure of an atom. </li></ul><ul><li>One device that can be used to determine the mass is a mass spectrometer. </li></ul><ul><li>One particular type is the Bainbridge Mass Spectrometer. </li></ul>
  19. 21. Bainbridge Mass Spectrometer <ul><li>Consists of three sections as shown below: </li></ul>
  20. 22. Bainbridge Mass Spectrometer <ul><li>Ion Source </li></ul><ul><ul><li>The gas is admitted into the chamber through a needle valve. </li></ul></ul><ul><ul><li>The gas discharge is established using a P.D. of 10,000V. </li></ul></ul><ul><ul><li>Electrons are stripped from the gas leaving the positive ions to move to the cathode. </li></ul></ul>
  21. 23. Bainbridge Mass Spectrometer <ul><li>Velocity filter: </li></ul><ul><ul><li>Ions pass through perpendicular uniform electric and magnetic fields. </li></ul></ul><ul><ul><li>They are attracted towards the negative plate due to the electric field with a force of: </li></ul></ul>
  22. 24. Bainbridge Mass Spectrometer <ul><li>They are also forced in the opposite direction due to the magnetic field with a force of: </li></ul><ul><li>If the net force is zero (F E = F B ), the ions will move through slit S 3 if: </li></ul><ul><li>E q = q vB </li></ul>
  23. 25. Bainbridge Mass Spectrometer <ul><li>The velocity of these ions will be: </li></ul><ul><li>By changing the magnetic and electric fields, ions with particular velocities can be selected. </li></ul>
  24. 26. Bainbridge Mass Spectrometer <ul><li>Magnetic deflection chamber </li></ul><ul><li>The chamber is evacuated to a very low pressure so that the ions can move without colliding with gas molecules. </li></ul><ul><li>As the force is always at right angles,to v and B ,so the ions move in a circular path. </li></ul>
  25. 27. Bainbridge Mass Spectrometer <ul><li>This path is described by the centripetal force equation: </li></ul><ul><li>So within the magnetic field: </li></ul>
  26. 28. Bainbridge Mass Spectrometer <ul><li>Rearranging the formula gives: </li></ul><ul><li>It is known that v , q and B are constant, therefore: </li></ul><ul><li>  r  m </li></ul>
  27. 29. Bainbridge Mass Spectrometer <ul><li>This will means ions of different mass will move in circles of different radii, and fall onto the photographic plate,as a series of lines. </li></ul>
  28. 30. Bainbridge Mass Spectrometer <ul><li>This device can determine the mass of various isotopes from the above equations and their relative amounts, by the relative intensities of the lines. </li></ul>
  29. 31. Bainbridge Mass Spectrometer <ul><li>If different atoms with masses m 1 and m 2; </li></ul><ul><ul><li>are introduced into the spectrometer, </li></ul></ul><ul><ul><li>the ratio for masses to radii is given by: </li></ul></ul><ul><li>Inaccuracies in measurement of quantities of B , v and q can be ignored a staggering seven-figure accuracy can be achieved. </li></ul>
  30. 32. Example <ul><li>a) A beam of chlorine-35 ions, each with a mass of 5.8 x 10 -26 kg, enters the evacuated chamber of a mass spectrograph with a velocity of 1.0 x 10 5 ms -1 . The magnetic induction in the chamber is 0.60 T. What is the radius of curvature of the beam? </li></ul><ul><li>b) If the beam contains some ions of chlorine-37, what will be the separation between the two beams at the deflector plate in the magnetic deflection chamber? </li></ul>
  31. 33. Solution – Part (a) <ul><li>B = 0.60 T </li></ul><ul><li>m= 5.8 x 10 -26 kg </li></ul><ul><li>v = 1.0 x 10 5 ms -1 </li></ul><ul><li>q = 1.6 x 10 -19 C </li></ul>
  32. 34. Solution – Part (a) <ul><li>In the magnetic deflection chamber, the centripetal force on the ions is supplied by the magnetic force: </li></ul><ul><li>F c = F B </li></ul>
  33. 35. Solution – Part (a) <ul><li>r = 0.0604 m </li></ul>
  34. 36. Solution – Part (b) <ul><li>m = 35 </li></ul><ul><li>r = 6.0 cm </li></ul><ul><li>M = 37 </li></ul><ul><li>R = ? </li></ul>
  35. 37. Solution – Part (b) <ul><li>As the charge of the ions in each beam is the same, and they are travelling through the same magnetic field, </li></ul>
  36. 38. Solution – Part (b) R = 6.4 x 10 -2 m
  37. 39. Solution – Part (b) <ul><li>The separation between the beams will be the difference in the diameters, for the two circular paths. </li></ul><ul><li>The diameter of the circle is twice the radius. </li></ul><ul><li>Separation = 2(6.39 – 6.04) </li></ul><ul><li>Separation = 7.0 x 10 -3 m </li></ul>
  38. 40. Nuclear Energy Levels <ul><li>Remember how e/m radiation is given off when electrons that have been excited, drop back down to lower energy levels. </li></ul><ul><li>The photons given off have specific energies equal to the difference in energy levels. </li></ul>
  39. 41. Nuclear Energy Levels <ul><li>Previously, we have studied  ,  , and  radiation </li></ul><ul><li>Often,  radiation has accompanied the emission of  or  particles. </li></ul><ul><li>The emission of  photons is evidence that the nucleus has energy levels </li></ul>
  40. 42. Nuclear Energy Levels <ul><li>Radium decays to Radon at different energy levels. </li></ul><ul><li> -particles are ejected at certain discrete velocities (energies). </li></ul><ul><li>The energy depends on which level the Radium decays to in the Radon. </li></ul>
  41. 43. Nuclear Energy Levels
  42. 44. Nuclear Energy Levels <ul><li>In the above diagram, 226 Ra decays giving off an  B particle that has a specific K (  B ) when it decays to Rn in the 2 nd excited state. </li></ul><ul><li>The 222 Rn then might move to the ground state giving off a photon of energy in the MeV range called a GAMMA PHOTON (  ) </li></ul>
  43. 45. Radioactive Decay <ul><li>We will look at the following areas: </li></ul><ul><ul><li>Beta Decay </li></ul></ul><ul><ul><li>The Neutrino </li></ul></ul><ul><ul><li>Half-life </li></ul></ul><ul><ul><li>Decay constant </li></ul></ul>
  44. 46. BETA DECAY <ul><li>Nuclei that have an imbalance of protons or neutrons can be unstable and also undergo radioactive decay. </li></ul><ul><li>The process involves the change of a proton into a neutron or more commonly a neutron into a proton with the ejection of an electron from the nucleus. </li></ul><ul><li>This decay is called beta decay, and the electron is referred to as a beta particle. </li></ul>
  45. 47. BETA + DECAY <ul><li>BETA + DECAY (too many protons)- </li></ul><ul><li>When a nucleus has to increase its neutron number to become more stable, a proton can spontaneously change into a neutron. </li></ul>B +
  46. 48. BETA + DECAY <ul><li>An electron (positively charged) is ejected with a neutrino. </li></ul><ul><li>The positive electron is called a positron and is an example of antimatter. The atomic number is reduced by one but the mass number is unaffected. </li></ul>
  47. 49. BETA + DECAY <ul><li>On the line stability on the graph, any atom below the line would decay this way. </li></ul>B +
  48. 50. BETA + DECAY <ul><li>In the nucleus, the reaction is: </li></ul><ul><li>An example of this is: </li></ul>
  49. 51. BETA + DECAY <ul><li>Notice that both mass and charge are conserved. </li></ul><ul><li>A ‘ positron ’, a positively charged electron (the same mass as an electron) is ejected. </li></ul><ul><li>The positron is an example of antimatter (“opposite of”). </li></ul><ul><li>This is known as ‘ proton decay ”. </li></ul>
  50. 52. BETA + DECAY <ul><li>The positron is known as the B + particle. </li></ul><ul><li>A Neutrino (v) is also released. </li></ul><ul><li>Note a new element is formed. There are no natural positron emitters since positron half-lives are very small. </li></ul><ul><li>Note - as the 13 N might decay into a metastable form of 13 C, the 13 C could then drop down to a more stable state, giving off a GAMMA RAY. </li></ul>
  51. 53. NEUTRINOS AND ANTINEUTRINOS <ul><li>Beta particles are emitted with a range of energies up to a maximum of a few MeV. </li></ul><ul><li>It seemed strange that the electrons with the maximum kinetic energy carried away all the available energy, yet those with less than he maximum kinetic energy appeared to have energy missing. </li></ul>
  52. 54. NEUTRINOS AND ANTINEUTRINOS <ul><li>This did not obey the law of conservation of energy. </li></ul><ul><li>Other experiments with momentum confirmed that linear momentum was not conserved . </li></ul>7 N 14 6 C 14 e - Speed and direction of the electron if momentum was conserved. e - Actual speed and direction of the electron.
  53. 55. NEUTRINOS AND ANTINEUTRINOS <ul><li>In 1934 Enrico Fermi developed the theory of beta decay and that the conservation laws did hold because there was a particle that had yet to be detected carrying the lost energy and momentum. </li></ul><ul><li>He called this particle a neutrino (Italian for ‘little neutral one’). </li></ul><ul><li>The antimatter of the neutrino ( ) is the antineutrino ( ). </li></ul>
  55. 57. NEUTRINOS AND ANTINEUTRINOS <ul><li>Using the conservation laws, he postulated the properties for the neutrino. </li></ul><ul><ul><li> Neutrinos are uncharged. This is because charge is already conserved. A neutron decays into a proton and an electron. </li></ul></ul><ul><ul><li> Neutrinos have zero rest mass but carry energy and momentum. The conservation laws would not hold otherwise. </li></ul></ul>
  56. 58. NEUTRINOS AND ANTINEUTRINOS <ul><ul><li> Neutrinos react very weakly with matter. It took 25 years to detect them and there are millions of neutrinos that pass through the Earth from the sun as if the Earth was not there. This is because they have no real mass or charge. </li></ul></ul><ul><ul><li> Neutrinos travel at the speed of light. As they have no mass but have energy, they must travel at the maximum speed possible - the speed of light. </li></ul></ul>
  57. 59. NEUTRINOS AND ANTINEUTRINOS <ul><li>The neutrino was accepted readily as it solved awkward problems but was not discovered until 1956. </li></ul><ul><li>It is given the symbol (the Greek letter nu) and has zero atomic number and mass number. </li></ul>
  58. 60. Radioactive Decay <ul><li>Radioactive decay is a completely random process. </li></ul><ul><li>No one can predict when a particular parent nucleus will decay into its daughter . </li></ul><ul><li>Statistics, however, allow us to predict the behaviour of large samples of radioactive isotopes. </li></ul>
  59. 61. Radioactive Decay <ul><li>We can define a constant for the decay of a particular isotope, which is called the half-life . </li></ul><ul><li>This is defined as the time it takes for the activity of the isotope to fall to half of its previous value. </li></ul>
  60. 62. Radioactive Decay <ul><li>From a nuclear point of view, the half-life of a radioisotope is the time it takes half of the atoms of that isotope in a given sample to decay. </li></ul><ul><li>The unit for activity, Becquerel (Bq), is the number of decays per second. </li></ul>
  61. 63. Radioactive Decay <ul><li>An example would be the half-life of tritium ( 3 H), which is 12.5 years. </li></ul><ul><li>For a 100g sample, there will be half left (50g) after 12.5 years. </li></ul><ul><li>After 25 years, one quarter (25g) will be left and after 37.5 years there will be one eighth (12.5g) and so on. </li></ul>
  62. 64. Radioactive Decay <ul><li>The decay curve is exponential.The only difference from one sample to another is the value for the half-life. </li></ul>
  63. 65. Radioactive Decay <ul><li>Below is a decay curve for 14 C. </li></ul><ul><li>Determine the half-life for 14 C. </li></ul>
  64. 66. Radioactive Decay <ul><li>The half-life does not indicate when a particular atom will decay but how many atoms will decay in a large sample. </li></ul><ul><li>Because of this, there will always be a ‘bumpy’ decay for small samples. </li></ul>
  65. 67. Radioactive Decay <ul><li>If a sample contains N radioactive nuclei, </li></ul><ul><ul><li>we can express the statistical nature of the decay rate </li></ul></ul><ul><ul><li>(- dN/dt ) </li></ul></ul><ul><ul><li>is proportional to N : </li></ul></ul>
  66. 68. Radioactive Decay <ul><li>in which  , the disintegration or decay constant , has a characteristic value for every radionuclide. This equation integrates to: </li></ul><ul><li>N o is the number of radioactive nuclei in a sample at t = 0 and N is the number remaining at any subsequent time t . </li></ul>You have to derive this
  67. 69. Radioactive Decay <ul><ul><li>- dN/dt =  N </li></ul></ul><ul><ul><li>Collect like terms </li></ul></ul><ul><ul><li>dN/N = -  dt </li></ul></ul><ul><ul><li>Integrate </li></ul></ul><ul><ul><li>ln N = -  t + c </li></ul></ul><ul><ul><li>But c = ln N 0 </li></ul></ul><ul><ul><li>So, ln N = -  t + ln N 0 </li></ul></ul><ul><ul><li>ln N - ln N 0 = -  t </li></ul></ul><ul><ul><li>N/ N 0 = e -  t </li></ul></ul><ul><li>Differentiating the radioactive decay law itself: </li></ul>
  68. 70. Radioactive Decay <ul><li>Solving for t ½ yields, that is when N = N 0 /2 </li></ul><ul><li> t 1/2 = 0.693 </li></ul><ul><li>t 1/2 = 0.693/  </li></ul>or
  69. 71. Radioactive Decay <ul><li>The half-life of an isotope can be determined by graphing the activity of a radioactive sample,over a period of time </li></ul>
  70. 72. Radioactive Decay <ul><li>The graph of activity vs time can be graphed in other ways </li></ul><ul><li>As the normal graph is exponential it does not lead to a straight line graph </li></ul><ul><li>Semi logarithmic graph paper can solve this problem </li></ul>
  71. 73. Radioactive Decay
  72. 74. Radioactive Decay <ul><li>If we take the natural log of N = N o e -  t we get: </li></ul><ul><li>ln N = ln N o -  t </li></ul><ul><li>The slope of the line will determine the decay constant  </li></ul>
  73. 75. Radioactive Decay <ul><li>The accuracy in determining the half-life depends on the number of disintegrations that occur per unit time. </li></ul><ul><li>Measuring the number of disintegrations for very long or short half-life isotopes could cause errors. </li></ul>
  74. 76. Radioactive Decay <ul><li>For very long half-life isotopes </li></ul><ul><ul><li>i.e. millions of years </li></ul></ul><ul><ul><li>Only a small number of events will take place over the period of a year </li></ul></ul><ul><li>Specific activity is used </li></ul><ul><li>Activity of sample is measured against a calibrated standard </li></ul>
  75. 77. Radioactive Decay <ul><li>Standard is produced by reputable organisations </li></ul><ul><ul><li>i.e. International Atomic Energy Agency </li></ul></ul><ul><li>Calibrated standard measures the accuracy of the detector making sure it is accurate </li></ul><ul><li>Specific activity and atomic mass of isotope is then used to calculate the half-life </li></ul>
  76. 78. Radioactive Decay <ul><li>With very short half-life isotopes, the isotope may disintegrate entirely before it is measured.Time is therefore of the essence </li></ul><ul><li>As most of these isotopes are artificial. Produce them in or near the detector </li></ul><ul><li>This eliminates or reduces the transfer time problem </li></ul>
  77. 79. EXAMPLE 1 <ul><ul><li>(a) Radium-226 has a half-life of 1622 years. A sample contains 25g of this radium isotope. How much will be left after 3244 years? </li></ul></ul><ul><ul><li>(b) How many half-lives will it take before the activity of the sample falls to below 1% of its initial activity? How many years is this? </li></ul></ul>
  78. 80. EXAMPLE 1 SOLUTION <ul><ul><li>(a) 3244 years is 2 half lives (2 x 1622) </li></ul></ul><ul><ul><li> N = N o (1/2) n </li></ul></ul><ul><ul><li>= 25 x (1/2) 2 </li></ul></ul><ul><ul><li>= 25 x (1/4) </li></ul></ul><ul><ul><li>= 6.25 </li></ul></ul>
  79. 81. EXAMPLE 1 SOLUTION <ul><ul><li>The activity of a radioactive sample is directly proportional to the number of remaining atoms of the isotope. After t 1/2 , the activity falls to ½ the initial activity. After 2 t 1/2 , the activity is ¼. It is not till 7 half-lives have elapsed that the activity is 1/128 th of the initial activity. </li></ul></ul><ul><ul><li>So, 7 x 1622 = 11354 years </li></ul></ul>
  80. 82. EXAMPLE 2 <ul><li>A Geiger counter is placed near a source of short lifetime radioactive atoms, and the detection count for 30-second intervals is determined. Plot the data on a graph, and use it to find the half-life of the isotope. </li></ul>
  81. 83. EXAMPLE 2 <ul><li>Interval Count </li></ul><ul><li>1. 12456 </li></ul><ul><li>2. 7804 </li></ul><ul><li>3. 5150 </li></ul><ul><li>4. 3034 </li></ul><ul><li>5. 2193 </li></ul><ul><li>6. 1278 </li></ul><ul><li>7. 730 </li></ul>
  82. 84. EXAMPLE 2 SOLUTION <ul><li>The data are plotted on a graph with the point placed at the end of the time interval since the count reaches this value after the full 30 seconds. </li></ul>
  83. 85. EXAMPLE 2 SOLUTION <ul><li>A line of best fit is drawn through the points, and the time is determined for a count rate of 12 000 in 30 seconds. Then the time is determined for a count rate of 6000, and 3000. </li></ul>
  84. 86. EXAMPLE 2 SOLUTION <ul><li>t (12 000) = 30s </li></ul><ul><li>t ( 6000) = 72s, so t 1/2 (1) = 42s </li></ul><ul><li>t ( 3000) = 120s, so t 1/2 (2) = 48s </li></ul><ul><li>  The time difference should have be the half-life of the sample. </li></ul>
  85. 87. EXAMPLE 2 SOLUTION <ul><li>Since we have two values, an average is taken. </li></ul>